Solve for $x$ and $y$ using elimination. $\begin{align*}8x+4y &= 1 \\ -8x+8y &= -6\end{align*}$
We can eliminate $x$ when its corresponding coefficients are negative inverses. Add the top and bottom equations. $12y = -5$ Divide both sides by $12$ and reduce as necessary. $y = -\dfrac{5}{12}$ Substitute $-\dfrac{5}{12}$ for $y$ in the top equation. $8x+4( -\dfrac{5}{12}) = 1$ $8x-\dfrac{5}{3} = 1$ $8x = \dfrac{8}{3}$ $x = \dfrac{1}{3}$ The solution is $\enspace x = \dfrac{1}{3}, \enspace y = -\dfrac{5}{12}$.